A sample of 64 cows is used to estimate the average weight of a herd’s population....

A sample of 64 cows is used to estimate the average weight of a herd’s population. If the population standard deviation is 96 pounds, what is the margin of error of the population average weight estimate for a 90% confidence interval?

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Expert Solution

Solution :

Given that,

Population standard deviation = = 96
Sample size = n =64

At 90% confidence level the z is

= 1 - 90% = 1 - 0.90 = 0.1

/ 2 = 0.1 / 2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

Margin of error = E = Z/2 * ( / $\sqrt$ n)

= 1.645 * ( 96 / $\sqrt$ 64 )

=19.74

orchestra answered 1 year ago

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